A 500 g air-track glider attached to a spring with spring constant 10 N/m is sitting at rest on a frictionless air track. A 250 g glider is pushed toward it from the far end of the track at a speed of 120 cm/s. It collides with and sticks to the 500 g glider. What are the amplitude and period of the subsequent oscillations?

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Hey, everyone in this problem, we have a block of mass, 450 g connected to a vertical wall by a spring having a spring constant of 20 newtons per meter. Initially, the 450 g mass is at rest in the equilibrium position. While another mass of 350 g is pushed towards that first block at a velocity of 90 centimeters per second causing the two masses to collide and stick together. They were told to assume no friction and were asked to determine the amplitude and period of the resulting oscillations. Now, we're given four answer choices A through D and they have a combination of two amplitudes, either 1.62 m or 0.8 m and two possible periods, either 1.3 seconds or 2.5 seconds. So let's go ahead and start with figuring out everything we've been told in this problem. And we've been told a lot of information, let's sort through all of it three kind of situations to consider. We have before the collision, we have immediately asked the collision. OK. And then we have some time later at this maximum displacement. OK. Or in other words, when we reach the amplitude of the aci toy motion, because we have these three kind of situations. And before the collision, we're told that we have two blocks. The first one has a mass which we'll call M one of 450 g. We want to convert this to our standard unit of kilograms and recall that to go from grams to kilograms, we divide by 1000. So this is gonna be 0.45 kg in this initial mass or this first mass M one. And we're gonna say that before the collision, we're gonna use a subscript B. So the velocity of this block is gonna be V one B. It's gonna be zero m per second because we're told that this block is at rest. All right. Now, the second block has a mass of 350 g. So M two is gonna be 350 g. We're to convert that just like we did M one by dividing by 1000 we get 0.35 kg in this block. The initial velocity of this block we're told is 90 centimeters per second. We want to convert this into meters per second. So we can multiply it by one m divided by centimeters because we know there are 100 centimeters in every meter. We're essentially dividing by 100 and we get 0. m per second. And let me write that actually down below. So it doesn't get mixed up with our next set of information. All right. Now we're gonna move to after the collision, ok. Before the collision, we're told that it, the blocks at equilibrium position. So this spring we have no, no spring force acting there. Um We don't need to worry about that after the collision, these blocks stick together. So the mass after the collision, which we're gonna call ma is gonna be the sum of the two blocks M one plus M two. This is gonna be 0.45 kg plus 0. kg, which gives us a mass of 0.8 kg. Now, we don't know the velocity after the collision. OK? We don't know how fast these are moving. We do know that the position or the displacement X A is gonna be zero m. OK? Immediately after that collision, the block is still at equilibrium position. And so the displacement is gonna be zero of that spring. Now we move on to some time later when we reach the maximum displacement or the amplitude of this a soul motion. OK? We're gonna call this the final time points. So we're gonna use subscript F. Now, the mass is gonna be the exact same as a mass after the collision. After those blocks stick together, they remain stuck together. And so the mass is still gonna be 0.8 kg, the velocity it's gonna be zero m per second. They recall that once you reach the maximum displacement, once you're at the amplitude, that block is gonna come to rest momentarily as it changes directions. And so that final velocity at the amplitude is gonna be zero m per second. And the position X F is gonna be the amplitude A. All right. So we have these three cases we have before and after the collision. And then at the maximum displacement, we wanna figure out what this amplitude A is. OK. In order to do that, we can use the conservation of mechanical energy. OK? From after the collision to this maximum displacement, but we don't know the velocity after the collision. OK. So in order to use our conservation mechanical energy to find the amplitude, we need this va value. How can we get that? Well, we have a collision, we can use our conservation of momentum. OK. And so we're gonna do just that and we're gonna look at the collision first. OK. So this is going from before collision to after the collision and we have a conservation of momentum. What that tells us is that the initial momentum P is gonna be equal to the final momentum P F. So initially we have two blocks. OK. So our initial momentum is made up of the momentum of each of those blocks. So we have P one B before the collision plus P two B is gonna be equal to P A, the momentum after the culture number call that momentum is given by the mass multiplied by the velocity. So for each of these terms, we're gonna get the mass multiplied by the corresponding velocity, we get M one V one B plus M two V two B is equal to ma va and this value va is what we are looking for right now that velocity after the collision so that we can go ahead and find our amplitude. Now we know that the initial velocity of block one is zero. So the first term on the left-hand side is gonna go to zero. We're left with M two which is 0.35 kg multiplied by its initial speed, 0.9 m per second. That's gonna equal that total mass of the blocks when they stick together after the collision of 0.8 kg multiplied by their velocity va. Now, in order to solve for VA, we are gonna go ahead and divide by 0. kg. The unit of kilogram will divide out. We're gonna be left with meters per second. OK. So we get that va it's gonna be equal to 0. 375 m per second. All right. So now we know the velocity after the collision and we can move on to using our conservation of mechanical energy to figure out the amplitude. Yeah. So going back to the information we were given, we used conservation of momentum to look at the collision. Now, we're gonna use conservation of mechanical energy to look at what happens immediately after the collision until this maximum displacement. So this is gonna be our post collision period. The conservation of mechanical energy tells us that the mechanical energy immediately after the collision E met A is gonna be equal to the mechanical energy at the end of this entire situation when we have this maximum amplitude and we're gonna call that E E F I recall that our mechanical energy is the sum of the kinetic energy and the potential energy. In this case, we have no gravitational potential energy. These blocks are on a surface, but we do have the potential energy due to the spring. OK. So for each, we're gonna have the potential energy and the kinetic energy we get K A plus plus E usa. OK. So the kinetic energy after the collision plus the potential energy due to the spring after the collision is equal to K F plus U F those same quantities at the very end when we have this should be us f sorry. And that's at the very end when we have that maximum amplitude. Now recall that the kinetic energy is given by one half M V squared, the potential energy for the spring is given by one half K X squared. And so our equation is going to become one half ma va squared plus one half K X A squared is equal to one half um V F squared plus one half K X F squared. All right. Now using the information that we have, we know that initially after the collision, we have no displacement X A is equal to zero. So the initial um potential energy to this spring is gonna be zero. OK. That term on the left-hand side and at the end when we have that maximum amplitude, we have a velocity of zero V F is equal to zero. So the first term on the right hand side is gonna go to zero as well. So we get one half ma va squared is equal to one half K X F squared. We can divide everything by one half. We're left with ma the mass after the collision, which is the total mass of the two blocks, 0.8 kg multiplied by their velocity, which we just found by using our conservation of momentum, 0.39375 m per second. OK? And this is equal to k the spring constant, which we're told is 20 newtons per meter multiplied by the amplitude squared. Simplifying on the left hand side, you get 0.124, 03125. OK? We're gonna use all the digits here just so we have no rounding error. Our units are kilogram meter per second. On the right side, we have 20 newtons per meter. Now recall that a Newton is equivalent to a kilogram meter per second squared. So if we take that divide it by meters, we're left with kilogram per second squared and that's multiplied by the amplitude squared. Now we're gonna divide by 20 kg per second squared to isolate for the amplitude. And we have, I have missed the square on the velocity, my apologies. So we have M V squared. So it should be 0.8 kg multiplied by 0.39375 m per second, all squared. OK. So then our units here on the left hand side are kilogram meter squared per second squared. All right, we're back on track. We're dividing by 20 kg per second squared. And we get that the amplitude squared A squared is equal to 0.0 m squared. And when we take this square route, OK? Remember that we're gonna get the positive and negative route. In this case, we're looking for the amplitude. That's just a distance. So we just want that positive value that's gonna be 0. m. OK? And writing out your units is really important. OK. That's how you can see that you've missed something like a square because the units don't match up. OK. We have an amplitude, we have the unit in meters, which is exactly what we want. All right. So we found the amplitude. Let's compare this to our answer choices, we can see that we can eliminate options A and B because they both have an amplitude of 1.62 m, which is not what we found. Answer choice C and D both have the amplitude of 0.8 m, which matches what we found if we round. And so we're left with finding the period now to determine if we have C or D. Now let's recall that the period T in simple harmonic motion is equal to two pi multiplied by the square root of M divided by K. So based on the information we were given in the problem, you can go ahead and calculate the period directly. We have two pi multiplied by the square root of the mass M. Now, when this is oscillating, that mass is the combined mass of those two blocks that are stuck together, which is 0.8 kg. And we divide that by the spring constant K, which is 20 newtons per meter. If we simplify inside the square root we have in the denominator Newton divided by meter that leaves us with a kilogram per second squared. So kilogram divided by kilogram per second squared is gonna give us the unit of second squared inside that square root, we get two pie multiplied by the square root of 0.4 second squared. So when we simplify this, using a calculator, we get a period of 1.26 seconds. OK. So that is that second part of the problem and we can not now go up and answer this problem. So we found an amplitude of approximately 0.8 m in a period of about 1.3 seconds which corresponds with answer choice. C Thanks everyone for watching. I hope this video helped see you in the next one.

Verified Solution

Key Concepts

Conservation of Momentum

Simple Harmonic Motion (SHM)

Amplitude and Period of Oscillation